I simulated the external wall and I would like to compare it with measurements data.The Thickness of layer and thermal conductivity of them are same but the total U-Value of the Measurement and simulated wall are not same.
How can be considered the indoor and outdoor surface thermal resistance?It should be defined new layer as Air or it has another solution?
Thank you so much
Indoor and out door surface thermal resistance
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Re: Indoor and out door surface thermal resistance
Hi samiraaien,samiraaien wrote:How can be considered the indoor and outdoor surface thermal resistance?It should be defined new layer as Air or it has another solution?
in the dialog "Surface Transfer Coeff." you can adjust the heat transfer coefficient for either surface (or, alternatively, the heat transfer resistance which is the reciprocal of the coefficient. Use the dialog "Options | Unit System" to switch between the coefficient and the resistance).
Regards,
Thomas
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Re: Indoor and out door surface thermal resistance
I simulated a wall with 5 layers and thermal resistance of indoor and outdoor surface have been considered.Also I calculated the R value and U value manually.It can be seen in attachment Picture that the result of this 2 value in simulation and manually calculation are not same while the number , thickness and thermal conductivity of layers and surface resistance are exactly same.
What is the problem and how can be reach to the calculation result via simulation?
What is the problem and how can be reach to the calculation result via simulation?
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- the picture shows the difference between result of R,U- Value in simulation and manually calculation.
- problem.PNG (276.37 KiB) Viewed 15747 times
Re: Indoor and out door surface thermal resistance
In order to determine a realistic U-value, WUFI uses the moisture-dependent heat conductivities (for 80% RH) of the materials for this evaluation, not the dry thermal conductivities.samiraaien wrote:What is the problem and how can be reach to the calculation result via simulation?
If a moisture-dependent thermal conductivity has been specified for a material (either given as a table, or generated from a moisture-dependent thermal conductivity supplement), WUFI uses the moisture storage function of this material to determine the water content at 80% RH, computes the thermal conductivity corresponding to this water content and uses the result to evaluate the U-value.
If a design value of the thermal conductivity has been specified for the material, this value will be used instead.
If you want to reproduce WUFI's result, you can either remove the moisture-dependence of the thermal conductivities of the involved materials, or you can follow the procedure described above to determine which thermal conductivity WUFI uses for each material and do your manual calculation with these values.
Regards,
Thomas
Re: Indoor and out door surface thermal resistance
Hi samiraaien,
here is a detailed calculation, reproducing WUFI's results. For each material, the equilibrium moisture at 80% RH is determined from the material's moisture storage function. Then the heat conductivity of the material for this water content is computed, either from the thermal conductivity supplement if the moisture-dependent heat conductivity is 'generated', or from interpolation in the respective table if the moisture-dependent heat conductivity is tabulated.
Cement Lime Plaster:
thickness, d = 0.002 m
water content at 80% RH, w(80%) = 45 kg/m3 (see moisture storage function)
heat conductivity at w(80%), lambda(45 kg/m3) = 0.47 * (1 + 8 * 45/1900) = 0.559 W/mK (see help topic on 'Heat Conductivity, Moisture-dependent')
Aspen Aerogels:
d = 0.04 m
w(80%) = 6.6 kg/m3
lambda(6.6 kg/m3) = 0.01487 W/mK (interpolated in table of moisture-dependent thermal conductivity)
Cement Lime Plaster:
d = 0.02 m
w(80%) = 45 kg/m3
lambda(45 kg/m3) = 0.8 * (1 + 8 * 45/1900) = 0.952 W/mK
Solid Brick:
d = 0.25 m
w(80%) = 11.8 kg/m3
lambda(11.8 kg/m3) = 0.445 W/mK (interpolated in table)
Gypsum Plaster:
d = 0.015 m
w(80%) = 1.77 kg/m3
lambda(1.77 kg/m3) = 0.201 W/mK (interpolated in table)
R = 0.002/0.559 + 0.04/0.01487 + 0.02/0.952 + 0.25/0.445 + 0.015/0.201 = 3.351 m2K/W, WUFI shows the R-value 3.35 m2K/W
U = 1/(0.0588 + 3.351 + 0.125) = 0.2829 W/m2K, WUFI shows the U-value 0.283 W/m2K
WUFI shows a total thickness of 0.33 m instead of the correct 0.327 m because the displayed total thickness is rounded to two decimal places; the current version shows three decimal places.
Also note, if you wish to change the thermal conductivity of a material:
In materials with tabulated moisture-dependent thermal conductivity it is not sufficient to change only the dry thermal conductivity. This leaves the rest of the table unchanged and if the water content of the material is non-zero (which is usually the case), the current thermal conductivity will always be taken from the unchanged part of the table.
You changed the dry thermal conductivity 0.014 W/mK of the aerogel to 0.029 W/mK, but you probably left the remaining table of moisture-dependent thermal conductivities unchanged, so for non-zero water contents your change has no effect (we are aware that this is a kind of 'hidden trap', and we are thinking about ways to make this more transparent to the user).
Kind regards,
Thomas
here is a detailed calculation, reproducing WUFI's results. For each material, the equilibrium moisture at 80% RH is determined from the material's moisture storage function. Then the heat conductivity of the material for this water content is computed, either from the thermal conductivity supplement if the moisture-dependent heat conductivity is 'generated', or from interpolation in the respective table if the moisture-dependent heat conductivity is tabulated.
Cement Lime Plaster:
thickness, d = 0.002 m
water content at 80% RH, w(80%) = 45 kg/m3 (see moisture storage function)
heat conductivity at w(80%), lambda(45 kg/m3) = 0.47 * (1 + 8 * 45/1900) = 0.559 W/mK (see help topic on 'Heat Conductivity, Moisture-dependent')
Aspen Aerogels:
d = 0.04 m
w(80%) = 6.6 kg/m3
lambda(6.6 kg/m3) = 0.01487 W/mK (interpolated in table of moisture-dependent thermal conductivity)
Cement Lime Plaster:
d = 0.02 m
w(80%) = 45 kg/m3
lambda(45 kg/m3) = 0.8 * (1 + 8 * 45/1900) = 0.952 W/mK
Solid Brick:
d = 0.25 m
w(80%) = 11.8 kg/m3
lambda(11.8 kg/m3) = 0.445 W/mK (interpolated in table)
Gypsum Plaster:
d = 0.015 m
w(80%) = 1.77 kg/m3
lambda(1.77 kg/m3) = 0.201 W/mK (interpolated in table)
R = 0.002/0.559 + 0.04/0.01487 + 0.02/0.952 + 0.25/0.445 + 0.015/0.201 = 3.351 m2K/W, WUFI shows the R-value 3.35 m2K/W
U = 1/(0.0588 + 3.351 + 0.125) = 0.2829 W/m2K, WUFI shows the U-value 0.283 W/m2K
WUFI shows a total thickness of 0.33 m instead of the correct 0.327 m because the displayed total thickness is rounded to two decimal places; the current version shows three decimal places.
Also note, if you wish to change the thermal conductivity of a material:
In materials with tabulated moisture-dependent thermal conductivity it is not sufficient to change only the dry thermal conductivity. This leaves the rest of the table unchanged and if the water content of the material is non-zero (which is usually the case), the current thermal conductivity will always be taken from the unchanged part of the table.
You changed the dry thermal conductivity 0.014 W/mK of the aerogel to 0.029 W/mK, but you probably left the remaining table of moisture-dependent thermal conductivities unchanged, so for non-zero water contents your change has no effect (we are aware that this is a kind of 'hidden trap', and we are thinking about ways to make this more transparent to the user).
Kind regards,
Thomas
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- WUFI User
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- Joined: Thu Dec 15, 2016 2:51 am -1100
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Re: Indoor and out door surface thermal resistance
Dear Thomas
Thank you so much for your quickly and completely answer.It was very useful for me.
Regards,
Samira
Thank you so much for your quickly and completely answer.It was very useful for me.
Regards,
Samira
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Re: Indoor and out door surface thermal resistance
Hello
I found this thread when I was looking for an answer to a related question:
Wufi takes as default the internal surface thermal resistance to be 0.125 m2.W/K. This is close to the value for walls specified in EN6946. Also, EN 13788 specifies that for assessing internal surface mould risk using the temperature factor method, a higher value of 0.25 should be used to represent the effect of furniture, still air near corners etc. I don't know where any of these values come from in terms of fundamental research. I know from our thermal bridge analysis work that the choice of internal surface resistance can have a significant effect on the mould risk, particularly for poorly insulated details or bad thermal bridges. So where internal mould is a risk, it seems this is a very important factor, but I can't find very much written about this.
Can anyone give any guidance on what values are appropriate for different situations, other than following the guidance in the standards mentioned above?
Thanks
Toby
I found this thread when I was looking for an answer to a related question:
Wufi takes as default the internal surface thermal resistance to be 0.125 m2.W/K. This is close to the value for walls specified in EN6946. Also, EN 13788 specifies that for assessing internal surface mould risk using the temperature factor method, a higher value of 0.25 should be used to represent the effect of furniture, still air near corners etc. I don't know where any of these values come from in terms of fundamental research. I know from our thermal bridge analysis work that the choice of internal surface resistance can have a significant effect on the mould risk, particularly for poorly insulated details or bad thermal bridges. So where internal mould is a risk, it seems this is a very important factor, but I can't find very much written about this.
Can anyone give any guidance on what values are appropriate for different situations, other than following the guidance in the standards mentioned above?
Thanks
Toby